$f(0)=-1\,$, $~f\,^\prime(0)=6\,$, $~f\,^{\prime\prime}(0)=\frac{2}{3}\,$, and $~f\,^{\prime\prime\prime}(0)=-6\,$. What are the first four nonzero terms of the Maclaurin series of $f$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $-1+6x+\frac{1}{3}{{x}^{2}}-{{x}^{3}}$ (Choice B) B $-x+6x^2+\frac{1}{3}{{x}^{3}}-{{x}^{4}}$ (Choice C) C $1+6x+\frac{1}{3}{{x}^{2}}-{{x}^{3}}$ (Choice D) D $-1+6x+\frac{2}{3}{{x}^{2}}-6{{x}^{3}}$ (Choice E) E $-1+6x+\frac{1}{3}{{x}^{2}}-2{{x}^{3}}$
Solution: We know the formula for a Taylor series centered at $~x=0\,$. $ f(0)+f\,^\prime(0)x+\frac{f\,^{\prime\prime}(0)}{2!}{{x}^{2}}+\frac{f\,^{\prime\prime\prime}(0)}{3!}{{x}^{3}}+...+\frac{{{f}^{(n)}}(0)}{n!}{{x}^{n}}+...$ If we substitute what is given for the first four values of the function and its derivatives, we get the following. $ T_3(x)=-1+6x+\frac{\frac{2}{3}}{2!}{{x}^{2}}-\frac{6}{3!}{{x}^{3}}$ This simplifies to $ T_3(x)=-1+6x+\frac{1}{3}{{x}^{2}}-{{x}^{3}}$.